UP TGT Mathematics Mock Test - 1 - UPTET MCQ

• Pair 1 is correctly matched: Non-violence (Ahimsa) is a core principle of Jainism, strongly associated with Mahavira.
• Pair 2 is correctly matched: The Eightfold Path is a fundamental teaching of Buddhism, introduced by Gautama Buddha as a means to end suffering.
• Pair 3 is incorrectly matched: Discarding clothes is associated with Mahavira and the Digambara sect of Jainism, not Chandragupta Maurya.

The Earth rotates from west to east. This means that when viewed from above the North Pole, the Earth rotates counterclockwise, causing the movement of the sun and stars from east to west in the sky.

• It was started by Sir Syed Ahmad Khan (1817-1898) for the Muslims' social and educational advancement in India.
• He fought against the medieval backwardness and advocated a rational approach towards religion. In 1866, he started the Muhammadan Educational Conference as a general forum for spreading liberal ideas among the Muslims.
• In 1875, he founded a modem school at Aligarh to promote English education among the Muslims. This had later grown into the Muhammadan Anglo-Oriental College and then into the Aligarh Muslim University.

The correct answer is Tappa.

Key Points

• Tappa is a form of Indian semi-classical vocal music that was derived from the folk music of Punjab and Sindh.
• Tappa is the oldest and most admired form of Pashto folk literature which originated from the folk songs of the camel riders in Punjab.
• It is one of the most difficult classical forms in Punjab.
• It was believed to be a song of the cameleers of Punjab and Rajasthan.

Additional Information

• Bhajan refers to any devotional song with a religious theme or spiritual ideas, specifically among Indian religions, in any of the languages from the Indian subcontinent.
• Qawwalis are the compositions and devotional music of Sufi saints who belong to the tradition of Islamic mysticism.
• Ghazal is a well-known poetic genre of Arabic literature.

CONCEPT:

Centre of gravity:

• The centre of gravity is a theoretical point in the body where the total weight of the body is thought to be concentrated.
• The centre of gravity of a body is a point where the total gravitational torque on the body is zero.
• The centre of gravity of a body may lie inside or outside the body.
• Let a body consists of n number of particles and the gravitational acceleration is uniform for all the particles of the body.
• Then the position of the centre of gravity is given as,

⇒ 

Where w1, w2,..., and wn = weight of the particles

EXPLANATION:

• The centre of gravity of the body coincides with the centre of mass in uniform gravity or gravity-free space.
• If the body is so extended that gravitational acceleration varies from part to part of the body, then the centre of gravity and centre of mass will not coincide.
• The centre of gravity of a plane lamina will not be at its geometrical centre if it is a Right angle triangle.

Additional Information

Centre of gravity of various plane areas

Given:

A is a square matrix of order 3 and |A| = -2

Concept used:

 = 

Calculations:

∵ AA^-1 = I

⇒ |AA^-1| = 1 

⇒ |A| |A^-1| = 1

∴ |A-1 |  = 

Given that , |A| = -2

∴ option 2 is correct 

Concept:

For two vectors  to be collinear,​  where λ is a scalar.

Order of a differential equation is defined as the order of the highest order derivative

 Degree of a differential equation, when it is a polynomial equation in derivatives, is defined as the highest power (positive integral index) of the highest order derivative.

The highest order derivative present in the differential equation is y’’’, so its order is three.

The given differential equation is not a polynomial equation in its derivatives and so its degree is 1

Concept:

Sum of first n natural numbers = n(n + 1)/2

Sum of cubes of first n natural numbers = 

Calculation:

Hence, option (2) is correct.

Option 4)

CONCEPT:

• Vector Product (Cross product): Its magnitude is equal to the products of the magnitude of two vectors and sine of the angle between them and whose direction is perpendicular to the plane of the two vectors.
  • î, ĵ, and k̂ are the unit vectors whose magnitude of each of the vector is 1
  • The angle betweenî, ĵ, and k̂ or any of the two of them is 90°. 

EXPLANATION:

• Vector product of orthogonal unit vectors, the magnitude of each of the vector î, ĵ, and k̂ is 1 or the angle between any of the two of them is 90° 

î × ĵ = (1)(1)sin90° n̂ = n̂ (sin90° = 1)

As n̂ is a unit vector perpendicular to the plane of î and ĵ, so it just the third vector k̂ 

î × ĵ = k̂, Similarly, we can write ĵ × k̂ = î, k̂ × î = ĵ, ĵ × î = -k̂, k̂ × ĵ = -i, î × k̂ = -ĵ

• The vector product of a vector with itself is a null vector. 

ĵ × ĵ = 0, k̂ × k̂ = 0, î × î = 0 

Additional Information

• Aid to memory
  •  Write î, ĵ, k̂ cyclically around a cycle, as shown in the figure.

Multiplying two-unit vector anticlockwise, we get a positive value of the third unit vector 

(î × ĵ = +k̂) and multiplying two-unit vector clockwise, we get the negative value of the third unit vector (ĵ × î = -î),

we can also conclude this from the cross product is anti- commutative that is A × B = - B × A

ĵ × k̂ = î, k̂ × î = ĵ, ĵ × î = -k̂, k̂ × ĵ = -i, î × k̂ = -ĵ 

Concept:

The general equation of any type of circle is represented by:

Where center of circle is given by (-g, -f)

Radius = 

If  Radius 1 + Radius 2 < d (Distance between the centers) then given two circles does not cut each other.

Calculation:

Given equation of circle 1 is  x2 + y2 - 6x - 14y + 48 = 0

and equation of circle 2 is x2 + y2 - 6x = 0

Now, Center of 1st circle:

2g = - 6 or g = - 3

2f = -14 0r f = - 7

Center of 1st circle = (3, 7)

Now, Center of 2nd circle:

2g = - 6 or g = - 3

2f = 0 or f = 0

Center of 1st circle = (3, 0)

Radius 1 + Radius 2 = 3 + √10 

And 3 + √10 < 7 (Distance between the centers)

∴ Circles are not touching each other.

∴ The number of common tangents = 4

Concept:

If two circles x² + y² + 2g₁x + 2f₁y + c₁ = 0 and x2 + y2 + 2g2x + 2f2y + c2 = 0 

are orthogonal then 2g₁g₂ + 2f₁f₂ = c₁ + c₂

Calculation:

If the circles x2 + y2 + 2x + 2ky + 6 = 0, x2 + y2 + 2ky + k = 0 intersect orthogonally,

⇒ 2(1)(0) + 2(k)(k) = 6 + k

⇒ 2k² - k - 6 = 0

⇒ (2k + 3)(k - 2) = 0

⇒ k = 2 or 

∴ The correct answer is option (1).

Given:

ABCD is a cyclic quadrilateral and  ∠B = 72°

Concept:

Cyclic is a quadrilateral whose all four vertices lie on the circle.

Also, the sum of opposite angles is equal to 180°.

Calculation:

ABCD is a cyclic quadrilateral

⇒ ∠B + ∠D = 180° ( Supplementary angles)

⇒ 72° + ∠D = 180°

⇒ ∠D = 108° 

Concept:

Limit of f(h) at h = a is said to exist if the function approaches the same value from both sides.

 f(h) = exist if LHL = RHL at h = a

Calculation:

We know that sin y lies between - 1 and 1.

⇒ - 1 ≤ sin y ≤ 1

Since  → ∞ 

⇒ - 1 ≤ sin  ≤ 1

LHL =  (0 - h)² sin $(0-\frac{1}{h})$

=   (h)2 sin  = 0

RHL =   (0 + h)2 sin 

=   (h)2 sin  = 0

⇒ LHL = RHL = 0

∴ The value of >

Concept:

Square matrix A is said to be skew-symmetric if a_ij = −a_ij for all i and j.

Square matrix A is said to be skew-symmetric if the transpose of matrix A is equal to the negative of matrix A ⇔ A^T = −A

All the main diagonal elements in the skew-symmetric matrix are zero.

Calculation:

For a skew-symmetric matrix, diagonal elements are zero and AT = −A

So, both and are Skew-symmetric matrix.

CONCEPT:

Centre of gravity:

• The centre of gravity is a theoretical point in the body where the total weight of the body is thought to be concentrated.
• The centre of gravity of a body is a point where the total gravitational torque on the body is zero.
• The centre of gravity of a body may lie inside or outside the body.
• Let a body consists of n number of particles and the gravitational acceleration is uniform for all the particles of the body.
• Then the position of the centre of gravity is given as,

⇒ $r=\frac{w_1{r}_1+w_2{r}_2+...+w_n{r}_n}{w_1+w_2+...+w_n}$

Where w1, w2,..., and wn = weight of the particles

EXPLANATION:

• The centre of gravity of the body coincides with the centre of mass in uniform gravity or gravity-free space.
• If the body is so extended that gravitational acceleration varies from part to part of the body, then the centre of gravity and centre of mass will not coincide. Hence, option 3 is correct.

Concept: Section formula is given by 

Calculation: 

Given: Position vector of A and B are  and  respectively

Let M be the mid-point of AB, such that AM = MB, it means that M divides AB in the ratio 1:1, so by section formula, we get,

 = 

Hence, we conclude that If the position vectors of A and B are  and  respectively, then the position vector of mid-point of AB is 

Given:

  for x ≠ 0 and f(0) = 0

Calculations:

Case 1 : if x > 0

⇒ 

Case 2 : if x < 0 

⇒ 

⇒ f(x) = 

At x = 0, we will check RHL and LHL

⇒ RHL =  

⇒ LHL = 

⇒ RHL = LHL = f(0) = 0

∴ f(x) is continuous everywhere.

Given:

PT = 96 cm

PB = 36 cm

OT = r

Concept used:

Here, ΔPTO is a right-angled triangle with ∠PTO = 90° as PT is the tangent on the quarter circle at point T though P, so using Pythagoras’ theorem

OP2 = OT2  + PT2 

Calculation:

According to the concept,

(OB + PB)² = OT2  + PT2 

⇒ (r + 36)2 = r2  + 962 

⇒ r2 + 36² + 72r = r2  + 962 

⇒ r = 110 cm

∴ Radius of the circle is 110 cm.

Concept:

• cos 2x = 2cos²x - 1 = 1 - 2sin²x
• 
• 1 + tan²x = sec²x
• ∫sec²ax dx = tan ax + C

∴ The correct answer is option (4).

Concept:

1. Equation of circle x2 + y2 = r2

2. Equation of an ellipse 

3. Equation of Parabola y2 = 4ax, x2 = 4ay

4. Equation of hyperbola 

5.If a2 = b2 then hyperbola is called rectangular hyperbola and x2 − y2 = a2 is the general form of a rectangular hyperbola

Calculation:

Given:

x = 3(cost + sint)

y = 4(cost - sint) 

 ----(1)

 ----(2)

Adding r=equation 1 and 2;

Given :

Equation of hyperbola is 5x2 - 4y2 = k2

Eccentricity(e) = 

Foci (± 2, 0)

Formula used :

General equation of hyperbola = 

Coordinates of foci (c) = ± ae           ------ (1)

Calculations :

⇒ 5x2 - 4y2 = k2

⇒ 

⇒ 

Comparing with general equation, we get

⇒ a² = k²/5 and b² = k²/4

Now we know that e = 3/2 and foci (± 2, 0)

Using equation (1), we get

⇒ c = ± ae

⇒ 

Formula used :

sin 2θ = 2 sin θ cos θ 

cos C + cos D = 2cos() cos()

Calculaton :

sin A - cos B - cos C = 0

⇒ sin A = cos B + cos C

Using the formula given above

⇒ 2 sin(A/2) cos(A/2) = 2cos() cos()

We know that, for a ΔABC, ∠ A + ∠ B + ∠ C = π 

⇒ sin(A/2) cos(A/2) = cos() cos()

Since, cos(π/2 - θ) = sin θ 

⇒ sin(A/2) cos(A/2) =  sin() cos()

⇒ cos(A/2) =  cos()

⇒ A/2 = (B - C)/2

⇒ A = B - C

⇒ B = A + C

But ∠ A + ∠ B + ∠ C = π 

⇒ B = π - B

⇒ 2B = π 

⇒  B = π/2

∴ Angle  B is equal to .

Concept:

• sin (A - B) = sin A cos B - cos A sin B
• The maximum value of a sin θ + b cos θ is 

Calculation:

CONCEPT:

• Projection of a vector  on other vector  is given by: 

CALCULATION:

Given:  and 

Here, we have to find the projection of a vector  on other vector  is given by: 

Concept:

If f(x) =  such that, 

,

{ form }

then, by L's hospital Rule,

Calculation:

∴ The correct answer is option (4).

Given:

Δ ABC is an equilateral triangle.

Formula Used:

Calculation:

From the given information the diagram can be drawn as

Let the side of an equilateral triangle be a cm

ΔABC is an equilateral triangle

BF ⊥ AC

⇒ AF = FC = a/2

Now,

sin 60° = EF/FC

tan 60° = EF/EC

Concept:

If 1, ω, ω² are the cube roots of unity,

•  and 
•       ----(1)

• 1 + ω + ω2 = 0 and ω³ = 1      ----(2)

Calculation:

We have (1 + ω2)(1 + ω4)(1 + ω8)(1 + ω16)

⇒ (1 + ω² + ω⁴ + ω⁶)(1 + ω¹⁶ + ω⁸ + ω²⁴)

⇒ (1 + ω2 + ω + 1)(1 + ω + ω² + 1)      [using (1)]

⇒ (0 + 1)(0 + 1)      [using (2)]

⇒ 1

Hence, (1 + ω2)(1 + ω4)(1 + ω8)(1 + ω16) = 1

Concept:

A differential equation is solved by separating the variables and then eliminating the differential terms by integrating both sides.

Calculation:

Given, 

⇒      -----(i)

Let y = v.x ⇒ 

Putting the value of y and  in (i),

⇒ 

∴ The correct answer is option (4).

Homomorphism:

Suppose the composition in G and G' are multiplicative. A mapping f from a group G into (onto) a group G' is said to be a homomorphism of G (onto) G' if,

⇒ f(xy) = f(x) f(y) ∀ x,y ∈ G.

f(G) is called homomorph or holomorphic image of G.

If f is homomorphism onto, then we say that G is holomorphic to G'. The symbol G^~ - G' is read as G is holomorphic to G'.